Let S= {1– (-1)" :n E N}. Then n2 3 1. inf(S) and sup(S)= 1 4 2. inf(S) = 1 and sup(S) = 2 3 3. inf(S) = and sup(S) = 2 4 %3D %3D 4. inf(S) = 0 and sup(S) = 1 5. None of these. Option 1 Option 2...

real analysis inf and sup

Extracted text: Let S= {1– (-1)" :n E N}. Then n2 3 1. inf(S) and sup(S)= 1 4 2. inf(S) = 1 and sup(S) = 2 3 3. inf(S) = and sup(S) = 2 4 %3D %3D 4. inf(S) = 0 and sup(S) = 1 5. None of these. Option 1 Option 2 Option 3 Option 4 Option 5